Hello. Solved problems are continuing. Preceding our problems resolved There were only x squared and y squared term. Or just had x y'l terms but There was two bi together. Gene continued with quadratic functions We will. Because in the 37th degree of application If you ever encounter the equation. Even if it is already off with bi shale important around certain critical points function is to understand how it behaves. For him, that's the second degree surfaces should know better. Now it is missing, we plus minus notion of residual plus We saw diminished or negative notion of residual terms these noble structures. x y'l did not. Bi x y'l just saw the plus or is missing. Now, in this example the y and x squared There squared term. Both have mixed x y'l terms. Now how these critical points We'll find? There are two methods for this, actually. With a derivative understand them. This analysis method. But there is a second method. This is an algebraic method completely. I also want to show it. Understanding this quadratic surface because Siya important b. First, let's start related to derivatives. You want to understand this surface. How to surface? Now to understand how to derivatives were we doing? Take the zero first order derivatives We mark. Critical points are important. How the single variable functions First we look at business critical points was happening. Here, too. Now if we take the derivative of it with respect to x. As you can see two x comes from here. Not a contribution here. Take the derivative with respect to x here by four years is coming. Similarly, if we take the derivative with respect to y Not a contribution here. There are two years here. Here comes the four x. Because x is a constant with a quartet doing the task. E is two equations for two unknown but The right side of the equation is zero which one to see if you can eliminate that, no Y'all, simplification of öbürkü scoop here will be zero. So the critical points still zero at zero. There is no change b. Now we ask the question before this one x We examine things squared plus y squared. x y'li we examined. This is how the surface of what will be? Full marks also the biggest plus to say Is it worth it, How Did the smallest value will be something critical surface? So easy to predict as we shall see can not be. And as always, plus valuable If we think maybe facing up will we tell paraboloid. We will see that it is wrong now. This intuitive prediction. Now to understand the nature of the critical points I need the second derivatives. As you can see here, if we take f x x two turns. If we take f x y four turns. Get here by the derivative with respect to y. Here again we take derivative with respect to x four involved. This is an internal consistency. we take the derivative with respect to y of the two involved. On these determinants If you see insert There are two on the diagonal as two. There are four four on the second diagonal. This is what you see when determinant account A four-minus ie less than zero minus twelve to sixteen. So when the determinant is less than zero What would be the minimum of what would be the maximum. So what what the minimum value maximum value happens. Only the saddle point would be. More thereof markings and We do not want to look at. Because this is a saddle point. Plus sign in it as you see it plus he marked off Shall we go one up, perhaps intuition as we are facing in the paraboloid we'd found. Here comes the opposite. What is the largest value of what the smallest. Now the second method is this:that this quadratic second moment function with it because it is algebraic operations that x We can not y'l term. When x y'l term not just the back prime structures that x squared and y squared terms remains. Then how the structure thereof is We immediately understood. Now we know the value of x at the moment A and B with X plus b c to d y y x plus y to say again to say c and d y do not know. If we place them we say these z function of x squared in there. There is a square plus y plus four times x times y is there. If x squared edit them their y squared between their among themselves and between them and x y'li Summing we get it. Our goal is the following:x to y will not. As you can see here if we do not x to y only the x and y squared squared terms will be. When this x squared and y squared in terms If the above signs plus it increases facing a paraboloid, if negative axis paraboloid or a downward one plus If one axis a saddle that is a hyperbolic Paraboloid We will see that point. Now here are four unknown. a b c d. Y against x to make them zero to zero olmasıda Our reset to force coefficient need. These four equivalent, one for the unknown We have the opportunity to choose three free equation. This is actually the subject matter of this course outside of The general subject of this investigation, but here a functional in terms of practical skills b would be useful to offer in terms of gain. For example, three of the four unknown seçsek as a a a, b, c au As you can see one one one he seçsek I'm sorry you two were here for two here here were four to six from two b c. Here c is a. or one. D turned six. So it is six plus six equals zero. So it is a negative going well. So we reset these terms. See also here we have chosen with a b c There was six. Where d is minus minus four here occurs. Here, the plus-plus-two occurs. So here is made minus two. Br br As you can see the sign plus sign minus. A saddle point is that we We know this basic principal structures for we have seen. This is a useful technique b. Not deep, but this is more than a b-side a beginning of a fundamental issue. This takes us to the matrix. Not our concern here. This is the third course in basic math. Linear algebra. If you recall the first four core courses We talked about the course. These are functions of one variable derivatives, even the infinite series of integral b. Or that's multivariate functions We examine the lesson. Also all of the terms B is linear multi-valued functions algebraic methods which, matrices that can be understood issues. This was our third lesson. The fourth lesson is the differential equations. Of course all of them here, we do not do. Only this relation would suffice to say. Here a little bit on the importance of linear algebra I'm highlighting concrete made. I'm giving you an assignment. BI like the previous. There are still x squared plus y squared term. The term BI has diminished. Wherein the term has a x y. This is how the surface? Place at a critical point and I want you to have the quality. A good exercise. After finding it again the question We can ask:is this what kind of surface? Maybe a minus point for being in the saddle This time we will suspect it might be. But at the beginning so easy to predict We see that. The answer turns out the smallest value. To do this your way clear. Take the second order derivatives of the determinant You will also place. Pros will appear. This plus after leaving the second with respect to x We will look for the sign of derivatives. From there, the smallest value such We will see that. In the second bi previous problems as we did instead of x with unknown numbers a and b is a algebraic transformation. an algebraic transformation instead of year again We do not use the term where x y'l To. Is there anything like that back then also. What a pit so that the surface is if you receive an upwardly facing side we see that parabolas. y you reset an upward parabola xi you reset upward facing a parabola. This is a hole in the surface. This is going elliptic paraboloid. Now a bit more complicated surfaces Let's see. How can we understand this surface? How can we draw? Wherein x has a term diced. cube has a term of years. There are also x y'l terms. This is now different from the surface of the second degree. But the method is always the same. Single accounts are a little thinner is here. To find the critical points with respect to x We take derivatives. Three x squared plus three years. We take the derivative with respect to y. From the first term does not contribute to y Get derivative is zero falls. Here frame of three years. x to y is a constant when receiving derivatives task for his three x remains. These two equations, two unknowns. As there is a fundamental difference from the previous. Here, the term now also consists squared. Now here come by for example y y minus x will be square. Here we put here to bring a little Edit your terms unless we obtain the following. Here y x y frame to frame is this time for the quaternary will be a term. This first order one quaternary a second degree more first-degree a force multiplied by the fourth degree function go to zero. Here's the real roots of the following quadratic terms You can immediately see that. Therefore, only x is equal to the real roots is zero and x is equal to minus one. x where x is equal to zero when When y is zero 0sıfırkoyun understood. z y x and y is zero when the zero understood to be. Alternatively x is equal to minus one See the when you get to x squared minus 1 plus about putting so y would be a negative one. y also be minus one. where x and y as a minus If you place minus two minus one minus one even more. But here is a plus minus times minus one for giving a Because it is three storeys plus a Z interests. So we find the critical points. As you can see here a previous of the problems Unlike several critical points print out. This is not something yadırgadıc. In a univariate function in You know, cubic more than the critical point of an equation I might. Now we get the second order derivatives f x were found. f x x're getting. We take f x and f y y y're getting. Now we have here, but two points In our critical point. These values are making zero zero. Then f x x is zero. f y o is zero. f x y three. f x y three. As you can see here, minus nine point value minus nine turns. Because it is less than zero determinant f x x is no longer do not look at or something to mark it a saddle I understand that point. When we come to the second floor minus x minus one year a. When we look here to x minus one When we put minus six. y minus x minus one we put in When this surplus. Now we insert it take x and y in time for three turns. There we found on the diagonal values. The best look of the diagonal large value or smallest value is we can see. Because the determinants plus valuable. However, for f x is x minus I understand that it was great value. Results compiled these results we zero zero A critical point was NOK. This saddle point. The other point is minus one minus one point becomes a point of maximum value. This is the greatest value to us he says. Now this is how we draw the surface? Now these surfaces to draw zero to zero in the Taylor expansion we look around on this one point giving value. It still around minus one minus one just as we did in terms of x y'l If we make a conversion to destroy As we arrive at a value. This surface will draw step by step. See a critical point of this surface zero minus one minus a critical point at zero on one. At this point. Now we have seen that a zero point zero There are saddle surface. E two sessions we saw the surface of x y'l prior to. Because the basic functions of the second degree I need to understand very well. So the point here a iöyl There is zero equal to zero and its contour value of such a thing lines. We've seen it around a saddle point how it behaves equivalent lines. At one point one of the largest value of the had. You bring the account, when you put here look in nature I need to draw them numerically much Not a benefit. If you want to put it on the computer You can remove. But it important to understand. We have such an initiative. Now it also follows on We can say. See where we have found that two critical Our point is. The maximum value of one point, one We have seen that the saddle point. Now here's the following year i.e., the first line is equal to x corner, judging by the angle symmetry revealed see that. y is equal to x equation As a cubic symmetry along line we obtain the equation. Something like this cubic equation. So I see progress this has a maximum value. From its maximum value such as I will now. Somewhere at the end will be zero. Here, the value becomes zero. Will be an inflection point. After that, he will grow. Now I started this line of equal value I draw here is such a saddle point there. Thereafter they continued if I There is now a critical value. In this attribute, such as qualitative continuing to be a growing thing. But combine them in any way you can like this lines to achieve something here. See here have such a hill. You can then coming down this hill. Such a landing then re- symmetry line is beginning to emerge in as it is on. Perhaps it may not be easy now to understand but at least to know this method When I need a little effort and in this way I need to understand. We were already doing in one variable? We're doing the same thing. We find the critical points thereafter curves there spend his quality I understand. There is a homework. Is a little different from the previous. Here cube, the cube had. Here are four forces. Here there were three times x and y. Here are four times x and y. However we can find the critical points. Here comes three critical points. Near the critical point of the Taylor sequences and where no complex terms obtained by We can. This series also from what they types of surfaces that As we can understand, and yet a surface can understand. Minus one one one one before critical points in a minus point We know that. This center point is a saddle We know. Means that roughly equal value such lines will occur. Here is the smallest value second derivatives to be understood derivatives taking. So we are moving in such a way and this intuitive Once you have completed the forms, but this closing so here goes. Would they see as eight figures is going on. Then around it such lines happens. This symmetry axis of symmetry we take on on the axis of the function easily you will find. Such would be a fourth order equation. See here from Great Value 're coming. Here you will come to zero. You'll leave here again again Are you down. See in this way the surface quality it is possible to understand. Thus a Madlab program, software is obtained by drawing. It also qualitatively We can understand. To draw on the computer, of course, already very gives accurate results but also to understand is this us so much in the analysis step by step to figure out what kind of surface is directs. Now here you are asked the following question:EUR now bivariate We have seen in the function, the critical values then how to find these critical values that the attributes of Does the minimum mA maximum saddle point does We understand that. What is the function of three variables, though? Even if the first three variable function order derivatives are zero. At the other, just f x, f y was zero. Here you will also f z is zero. There are three unknown. There are also three equations. This is a critical point of the solution of equations found. Or critical points. But with this knowledge sufficient conditions we can not do. Here again, as in a previous following matrix is obtained. On the diagonal of x x y z z'li. As you can see here two binary f x There were x y. Here, too, there were mixed derivatives. Six in number, but the mixed derivatives z'l one of x, y is z'l b x y'l as is also such an old Remove the three by three matrix. The determinant of this matrix and properties that tells us what kind of surface. Thus, more than two variables understand need for knowledge of the matrix. B This matrix has a special name. Or hessian matrix Hess said. But to do this basic The third lesson of mathematics, linear algebra I need to read. Although it will be many transactions seems to be the same. But this matrix to examine the two As in the binary is not easy. There was enough to get only determinant. I need some work here, but this our but knowledge of the subject goes beyond I need these can be examined. I'm giving you an assignment. This assignment is not symmetric as well öbürkü. There Küplü term, have squared term, of z'l There are terms that frame. That is a function of three unknowns. A function of three variables. x has the term even linear terms y'l There are x and y. Providing business critical point to be made here equations. Here is a minus x squared plus y is happening. We find f y. for we find z. We have three unknown x y z. The three were chosen so that a little unknowable three also found in all of them in a moment equations not kind. There wherein a linear equation. So that can be taken in account for example, now here is equal to y understood. z y not put in right here b y of X, the equation turns out. In a linear equation. If you find the place where y here involved an equation in terms of x. Therefore, not as unsolvable. Here you can understand easily, To watch the things you do results was also made in the accounts controlling accuracy of curvature he can. But this critical point, the nature of not our job to determine. This matrix algebra, linear algebra without work can be done from the but at least the critical points We find their place. Taylor series around these points opening Behavior near these spots can understand. Bye for now. Reviewing this issue until we meet again I would advise you to spend. Not very difficult issues. Accounts, particularly under the We choose unmanageable complexity In the examples you both need parsed homework I left as in the problem. After that we will deal with integration. Until then, your goodbye.